Page No 204:

Question 1:

Answer:

Page No 204:

Question 2:

(ii) A
line intersecting a circle in two points is called a __________.

(iii) A
circle can have __________ parallel tangents at the most.

(iv) The
common point of a tangent to a circle and the circle is called ____.

Answer:

(i) One

(ii) Secant

(iii) Two

(iv) Point
of contact

Page No 204:

Question 3:

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm. (B) 13 cm (C) 8.5 cm (D) cm test

Answer:

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

OP PQ

By applying Pythagoras theorem in ΔOPQ,

OP2 + PQ2 = OQ2

52 + PQ2 =122

PQ2 =144 − 25

PQ = cm.

Hence, the correct answer is (D).

Page No 204:

Question 4:

Draw
a circle and two lines parallel to a given line such that one is a
tangent and the other, a secant to the circle.

Answer:

It
can be observed that AB and CD are two parallel lines. Line AB is
intersecting the circle at exactly two points, P and Q. Therefore,
line AB is the secant of this circle. Since line CD is intersecting
the circle at exactly one point, R, line CD is the tangent to the
circle.

Page No 209:

Answer:

Page No 213:

Question 1:

From a point Q, the
length of the tangent to a circle is 24 cm and the distance of Q from
the centre is 25 cm. The radius of the circle is

(A) 7 cm (B) 12 cm

(C) 15 cm (D) 24.5 cm

Answer:

Let O be the centre of
the circle.

Given that,

OQ = 25cm and PQ = 24
cm

As the radius is
perpendicular to the tangent at the point of contact,

Therefore, OP ⊥
PQ

Applying Pythagoras
theorem in ΔOPQ, we obtain

OP2 + PQ2
= OQ2

OP2 + 242
= 252

OP2 = 625 −
576

OP2 = 49

OP = 7

Therefore, the radius
of the circle is 7 cm.

Hence, alternative (A)
is correct.

Page No 213:

Question 2:

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110, then ∠PTQ is equal to

(A) 60 (B) 70

(C) 80 (D) 90

Answer:

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90º

∠OQT = 90º

In quadrilateral POQT,

Sum of all interior angles = 360

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360

⇒ 90+ 110º + 90 +PTQ = 360

⇒ PTQ = 70

Hence, alternative (B) is correct.

Page No 213:

Question 3:

If tangents PA and PB
from a point P to a circle with centre O are inclined to each other
an angle of 80,
then ∠POA is equal to

(A) 50 (B)
60

(C) 70 (D)
80

Answer:

It is given that PA and
PB are tangents.

Therefore, the radius
drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥
PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior
angles = 360

∠OAP + ∠APB
+∠PBO + ∠BOA
= 360

90
+ 80
+90º +BOA
= 360

∠BOA = 100

In ΔOPB
and ΔOPA,

AP = BP (Tangents
from a point)

OA = OB (Radii of the
circle)

OP = OP (Common side)

Therefore, ΔOPB
≅ ΔOPA (SSS
congruence criterion)

A ↔
B, P ↔ P, O ↔
O

And thus, ∠POB
= ∠POA

Hence, alternative (A)
is correct.

Page No 214:

Question 4:

Prove that the tangents
drawn at the ends of a diameter of a circle are parallel.

Answer:

Let
AB be a diameter of the circle. Two tangents PQ and RS are drawn at
points A and B respectively.

Radius
drawn to these tangents will be perpendicular to the tangents.

Thus,
OA ⊥ RS and OB ⊥
PQ

∠OAR
= 90º

∠OAS
= 90º

∠OBP
= 90º

∠OBQ
= 90º

It
can be observed that

∠OAR
= ∠OBQ (Alternate interior
angles)

∠OAS
= ∠OBP (Alternate
interior angles)

Since
alternate interior angles are equal, lines PQ and RS will be
parallel.

Page No 214:

Question 5:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,

∠O’PB = 90° … (1)

O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠OPB = 90° … (2)

Comparing equations (1) and (2), we obtain

∠O’PB = ∠OPB … (3)

From the figure, it can be observed that,

∠O’PB < ∠OPB … (4)

Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

Page No 214:

Question 6:

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:

Let us consider a circle centered at point O.

AB is a tangent drawn on this circle from point A.

Given that,

OA = 5cm and AB = 4 cm

In ΔABO,

OB ⊥ AB (Radius ⊥ tangent at the point of contact)

Applying Pythagoras theorem in ΔABO, we obtain

AB2 + BO2 = OA2

42 + BO2 = 52

16 + BO2 = 25

BO2 = 9

BO = 3

Hence, the radius of the circle is 3 cm.

Page No 214:

Question 7:

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⊥ PQ (As OA is the radius of the circle)

Applying Pythagoras theorem in ΔOAP, we obtain

OA2 + AP2 = OP2

32 + AP2 = 52

9 + AP2 = 25

AP2 = 16

AP = 4

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord)

PQ = 2AP = 2 × 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.
[[VIDEO:14288]]

Page No 214:

Question 9:

In the given figure, XY
and X’Y’ are two parallel tangents to a circle with
centre O and another tangent AB with point of contact C intersecting
XY at A and X’Y’ at B. Prove that ∠AOB=90.

Answer:

Let us join point O to
C.

In ΔOPA and ΔOCA,

OP = OC (Radii of
the same circle)

AP = AC (Tangents
from point A)

AO = AO (Common side)

ΔOPA
ΔOCA
(SSS congruence criterion)

Therefore, P ↔
C, A ↔ A, O ↔
O

∠POA = ∠COA …
(i)

Similarly, ΔOQB
ΔOCB

∠QOB = ∠COB
… (ii)

Since POQ is a diameter
of the circle, it is a straight line.

Therefore, ∠POA
+ ∠COA + ∠COB
+ ∠QOB = 180º

From equations (i)
and (ii), it can be observed that

2∠COA
+ 2 ∠COB = 180º

∠COA + ∠COB
= 90º

∠AOB = 90°

Page No 214:

Question 10:

Prove that the angle
between the two tangents drawn from an external point to a circle is
supplementary to the angle subtended by the line-segment joining the
points of contact at the centre.

Answer:

Let us consider a
circle centered at point O. Let P be an external point from which two
tangents PA and PB are drawn to the circle which are touching the
circle at point A and B respectively and AB is the line segment,
joining point of contacts A and B together such that it subtends ∠AOB
at center O of the circle.

It can be observed that

OA (radius) ⊥
PA (tangent)

Therefore, ∠OAP
= 90°

Similarly, OB (radius)
⊥ PB (tangent)

∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior
angles = 360º

∠OAP +∠APB+∠PBO
+∠BOA = 360º

90º + ∠APB
+ 90º + ∠BOA = 360º

∠APB + ∠BOA
= 180º

Hence, it can be
observed that the angle between the two tangents drawn from an
external point to a circle is supplementary to the angle subtended by
the line-segment joining the points of contact at the centre.

Page No 214:

Question 11:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Since ABCD is a parallelogram,

AB = CD …(1)

BC = AD …(2)

It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.
[[VIDEO:14290]]

Page No 214:

Question 12:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In ABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = x (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Either x+14 = 0 or x − 7 =0

Therefore, x = −14and 7

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

Page No 214:

Question 13:

Prove that opposite
sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle.

Answer:

Let ABCD be a
quadrilateral circumscribing a circle centered at O such that it
touches the circle at point P, Q, R, S. Let us join the vertices of
the quadrilateral ABCD to the center of the circle.

Consider ΔOAP
and ΔOAS,

AP = AS (Tangents
from the same point)

OP = OS (Radii of the
same circle)

OA = OA (Common
side)

ΔOAP ≅
ΔOAS (SSS congruence
criterion)

Therefore, A ↔
A, P ↔ S, O ↔
O

And thus, ∠POA
= ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2
+ ∠3 + ∠4
+ ∠5 + ∠6
+ ∠7 + ∠8
= 360º

(∠1
+ ∠8) + (∠2
+ ∠3) + (∠4
+ ∠5) + (∠6
+ ∠7) = 360º

2∠1
+ 2∠2 + 2∠5
+ 2∠6 = 360º

2(∠1
+ ∠2) + 2(∠5
+ ∠6) = 360º

(∠1
+ ∠2) + (∠5
+ ∠6) = 180º

∠AOB + ∠COD
= 180º

Similarly, we can prove
that ∠BOC + ∠DOA
= 180º

Hence, opposite sides
of a quadrilateral circumscribing a circle subtend supplementary
angles at the centre of the circle.